Simplify; express your answer in exponential form. Assume $y\neq 0, t\neq 0$. $\dfrac{{(y^{3})^{3}}}{{(y^{-3}t^{-4})^{-4}}}$
Explanation: To start, try working on the numerator and the denominator independently. In the numerator, we have ${y^{3}}$ to the exponent ${3}$ . Now ${3 \times 3 = 9}$ , so ${(y^{3})^{3} = y^{9}}$ In the denominator, we can use the distributive property of exponents. ${(y^{-3}t^{-4})^{-4} = (y^{-3})^{-4}(t^{-4})^{-4}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(y^{3})^{3}}}{{(y^{-3}t^{-4})^{-4}}} = \dfrac{{y^{9}}}{{y^{12}t^{16}}}$ Break up the equation by variable and simplify. $\dfrac{{y^{9}}}{{y^{12}t^{16}}} = \dfrac{{y^{9}}}{{y^{12}}} \cdot \dfrac{{1}}{{t^{16}}} = y^{{9} - {12}} \cdot t^{- {16}} = y^{-3}t^{-16}$.